CBSE Sample Paper 2022-23 · Section C · 3 marks

Q27: Coin tossing for the odd person, or the mean number of defective items

Sample 2022-23Ch 13 · Probability

The question

(a) Three friends go for coffee. They decide who will pay the bill, by each tossing a coin and then letting the “odd person” pay. There is no odd person if all three tosses produce the same result. If there is no odd person in the first round, they make a second round of tosses and they continue to do so until there is an odd person. What is the probability that exactly three rounds of tosses are made?

(b) Find the mean number of defective items in a sample of two items drawn one-by-one without replacement from an urn containing 6 items, which include 2 defective items. Assume that the items are identical in shape and size.

Part (b) uses a random variable and its mean. This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

(a) In a single round there is no odd person when all three of them throw heads or all three throw tails.

\begin{aligned} &P(\text{no odd person in a round}) \\ &= \frac12 \times \frac12 \times \frac12 \times 2 = \frac14 \end{aligned}

Why?

Each coin shows a head or a tail with probability $\frac12$ ⁠, and the three tosses are independent, so we multiply:

\begin{aligned} P(HHH) &= \frac12 \times \frac12 \times \frac12 = \frac18 \\ P(TTT) &= \frac18 \end{aligned}

These two cannot happen together, so their probabilities add: $\frac18 + \frac18 = \frac14$ ⁠. That is where the $\times 2$ comes from.

\begin{aligned} &P(\text{odd person in a round}) \\ &= 1 - \frac14 = \frac34 \end{aligned}

Why?

In every round there either is an odd person or there is not, so these are complementary events:

P(\overline{E}) = 1 - P(E)

\begin{aligned} 1 - \frac14 &= \frac44 - \frac14 \\ &= \frac{4 - 1}{4} = \frac34 \end{aligned}

Exactly three rounds are made when there is no odd person in the first round and in the second round, and there is an odd person in the third round. The rounds are independent, so

\begin{aligned} &\text{Required probability} \\ &= \frac14 \times \frac14 \times \frac34 = \frac{3}{64} \end{aligned}

Why?

The game stops at the first round with an odd person. So exactly three rounds means: fail, fail, then succeed. Separate rounds do not affect each other, so the three probabilities multiply.

\begin{aligned} \frac14 \times \frac14 \times \frac34 &= \frac{1 \times 1 \times 3}{4 \times 4 \times 4} \\ &= \frac{3}{64} \end{aligned}

Watch out: the answer is not $\left(\frac14\right)^3$ ⁠. The third round must produce an odd person, otherwise a fourth round is made.

$\therefore$ The probability that exactly three rounds are made is $\frac{3}{64}$ ⁠.

(b) Let $X$ be the number of defective items in the sample. Then $X$ can take the values 0, 1 and 2.

\begin{aligned} P(X = 0) &= \frac46 \times \frac35 = \frac25 \\ P(X = 1) &= 2 \times \left(\frac26 \times \frac45\right) = \frac{8}{15} \\ P(X = 2) &= \frac26 \times \frac15 = \frac{1}{15} \end{aligned}

Why?

The urn has 2 defective and 4 good items. The draws are without replacement, so after the first draw only 5 items are left.

$X = 0$ ⁠: good, then good, so $\frac46 \times \frac35$ ⁠.

\begin{aligned} \frac46 \times \frac35 &= \frac{4 \times 3}{6 \times 5} = \frac{12}{30} \\ &= \frac{12 \div 6}{30 \div 6} = \frac25 \end{aligned}

$X = 1$ ⁠: defective then good, or good then defective. Both orders give the same product, which is why there is a factor 2:

\frac26 \times \frac45 = \frac46 \times \frac25 = \frac{8}{30}

\begin{aligned} 2 \times \frac{8}{30} &= \frac{16}{30} \\ &= \frac{16 \div 2}{30 \div 2} = \frac{8}{15} \end{aligned}

$X = 2$ ⁠: defective, then the one defective left out of 5.

\begin{aligned} \frac26 \times \frac15 &= \frac{2}{30} \\ &= \frac{2 \div 2}{30 \div 2} = \frac{1}{15} \end{aligned}

Check: $\frac{6}{15} + \frac{8}{15} + \frac{1}{15} = 1$ ⁠.

\begin{aligned} \frac25 &= \frac{2 \times 3}{5 \times 3} = \frac{6}{15} \\ 6 + 8 + 1 &= 15,\ \frac{15}{15} = 1 \end{aligned}

\begin{array}{c|c|c|c} x_i & 0 & 1 & 2 \\ \hline p_i & \frac25 & \frac{8}{15} & \frac{1}{15} \\ \hline p_ix_i & 0 & \frac{8}{15} & \frac{2}{15} \end{array}

\text{Mean} = \sum p_ix_i = \frac{8}{15} + \frac{2}{15} = \frac{10}{15}

Why?

The mean (expectation) of a random variable is the sum of each value times its probability:

E(X) = \sum p_ix_i

The value 0 adds nothing. Finally $\frac{10}{15} = \frac23$ ⁠, dividing top and bottom by 5.

\begin{aligned} 0 \times \frac25 &= 0 \\ 1 \times \frac{8}{15} &= \frac{8}{15} \\ 2 \times \frac{1}{15} &= \frac{2}{15} \\[4pt] \frac{8}{15} + \frac{2}{15} &= \frac{10}{15} \\ &= \frac{10 \div 5}{15 \div 5} = \frac23 \end{aligned}

$\therefore$ Mean number of defective items $= \dfrac23$

Where this comes from

Chapter 13, Probability: Independent trials; mean of a random variable.

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